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re: utility costs
9 apr 1999
marcia  wrote:
 
>investigate passive solar. it doesn't involve the huge expense you are
>referring to. it is a very simple idea that has been around for centuries.

direct gain, aka "mass and glass" (eg a massy floor in front of south
windows between the 24 hour living space and the outdoors) seems a little
too simple, at this point. performance can be improved a lot with some
changes, in many parts of the country.

>with passive solar you aren't adding any expensive active solar panels, etc. 

and sunspaces can serve other purposes, which improves the economics.
an add-on sunspace with an insulated wall between it and the house
(eg the original south wall of the house) and very little thermal mass
can be an efficient house heater. adding lots of thermal mass to a
sunspace as the passive solar industry council (ie the national concrete
masonry association and the brick institute of america) recommend raises
its price and cripples its performance...

>you utilize the sun with a heat sink to retain the solar energy gathered
>during the day. 

(a "heat sink" is a cooling device with no memory, vs a heat store, no?)

that sounds like direct gain again, aka "direct loss." lots of heat leaves
the house at night and on cloudy days through the solar glazing; you have
to live inside the heat store with its temperature fluctuations; it can't
be too warm for people and it's basically uninsulated, so it can't stay warm
for more than 1 cloudy day or so; too little glass means too little solar
gain, while too much means too much cloudy day heat loss, and so on.

>many homes are completely run on these simply systems in all parts
>of the country. 

it isn't hard to 100%-solar-heat a house. just remove the heating system,
leaving only the simply system. some people define "solar houses" as "those
with no other form of heat," which seems more inherently honest than talking
about "solar houses" that are only 30% solar heated. the issues then become
comfort, eg temperature extremes, and initial and operating cost. i doubt
there are any comfortable direct gain homes with no other form of heat in
seattle or anchorage or binghamton or detroit...

how can we improve the performance of a direct gain ("passive solar") home
by changing the configuration of mass and glass? where i live, the average
temperature in january is about 30 f, and an average of 1,000 btu per square 
foot of solar energy per day falls on a south wall, so a 1' "direct gain"
cube with an r1 south window with 100% solar transmission and r5 sides has
a thermal conductance of 1ft^2/r1 = 1 btu/h-f for the window and 5ft^2/r5 
= 1 for the sides, a total u = 2 btu/h-f. if the solar energy that flows
into the cube on an average day equals the heat energy that flows out, and 
the cube contains lots of thermal mass, so the room temperature t changes
slowly, then 1000 btu = 24h(t-30f)2btu;/h-f, and t = 50.8 f. coolish.

now suppose the cube only contains c = 50 btu/f of thermal capacitance,
like 50 pounds of water or 300 pounds of concrete (which won't even fit
in the box--water stores about 3x more heat than masonry by volume, and
it can be less expensive.) then rc = c/u = 25 hours, and the room temp
after 3 cloudy 30 f days in a row is 30+(50.8-30)exp(-72h/25h) = 31.2.
(well, 32, as the water freezes.) very coolish.

let's add another r5 wall behind the window to make a shallow low-thermal-
mass sunspace, and a low-power fan with a thermostat that turns on the fan
when the sunspace is warmer than the house. say the fan runs for 6 hours
on an average january day. then we have

   1000 btu = 6h(t-30)1ft^2/r1     for the sunspace, during the day
           + 18h(t-30)1ft^2/r6     for the sunspace, at night
           + 24h(t-30)5ft^2/r5     for the other cube walls,

            = (6+3+24)(t-30) = 33(t-30),

       so t = 30+1000/33 = 60.3 f, almost 10 degrees warmer... 

this configuration has a thermal conductance of 1ft^2/r6 + 5ft^2/r5 =
1.17 btu/h-f on a cloudy day, so rc = 50/1.17 = 43 hours, and the room
temp after 3 cloudy 30 f days is 30+(60.3-30)exp(-72/43) = 35.6. at
least the pipes don't freeze.

let's try more solar glazing. suppose this "house" is twice as long in
the ew direction (a "solar orientation.") the direct gain version then
has a conductance of 2 for the window and 8ft^2/r5 = 1.6 for the walls,
so 2000 = 24(t-30)3.6 and the average room temp t = 53.1. a little better,
but rc decreases to 50/3.6 = 13.9 hours, and after 3 cloudy 30 f days,
t = 30+(53.1-30)exp(-72/13.9) = 30.1. brrrr.

with the low-thermal-mass sunspace, 

   2000 btu = 6h(t-30)2ft^2/r1     for the sunspace, during the day
           + 18h(t-30)2ft^2/r6     for the sunspace, at night
           + 24h(t-30)8ft^2/r5     for the other cube walls,

            = (12+6+38.4)(t-30) = 56.4(t-30),

       so t = 30+2000/56.4 = 65.5 f, 

and the cloudy day conductance is 2ft^2/r6 + 8ft^2/r5 = 1.93 btu/h-f,
so rc = 50/1.93 = 26 hours, so the room temp after 3 cloudy 30 f days
is 30+(65.5-30)exp(-72/26) = 32.2. brrr. 

let's try an 8' cube with r8 insulation. the heat loss increases with
the square of the edge length, but the volume increases with the cube,
so we can stuff more thermal mass inside or use a smaller percentage
of floorspace with a larger cube to keep it warm in cloudy weather.

the direct gain version (in the brown colors) has a 64 btu/h-f for the
window and 5x64ft^2/r8 = 40 for the walls, a total of 104, so 64k btu
= 24h(t-30)104, and t = 55.6, after a long string of average days.
(increasing the wall insulation helped.) if this cube contains just
8 times more water (scaled up from a 1' cube, it would contain 256 times
more), ie 400 pounds, rc = 400/104 = 3.8 hours, and after 3 cloudy days,
t = 30+(55.6-30)exp(-72/3.8) = 30.0000002 f. brrrrr. (well, maybe 32 f,
for a few more days, until the house entirely freezes.) 

the low-thermal-mass sunspace version (in the green colors) has

    64k btu = 6h(t-30)64ft^2/r1     for the sunspace, during the day
           + 18h(t-30)64ft^2/r9     for the sunspace, at night
           + 24h(t-30)5x64ft^2/r8   for the other cube walls,

            = (384+128+960)(t-30) = 1472(t-30),

       so t = 30+64k/1472 = 73.5 f :-) 

and the cloudy day conductance is 64ft^2/r9 + 5x64ft^2/r8 = 47.1 btu/h-f,
so rc = 400/47.1 = 8.5 hours, so the room temp after 3 cloudy 30 f days
is 30+(65.5-30)exp(-72/8.5) = 30.009. brrr. 

it looks like the sunspace house is winning the steady-state performance
race, but cloudy-day performance is still a problem. what can we do?
not much, for the direct gain house, with all that window heat loss...

but the sunspace house might have a solar closet. suppose we put that same
400 pounds of water in its own compact 2' cube completely surrounded by r8
insulation, with its own 2x2' r1 glazing over the south wall insulation,
facing the sunspace, like this: 

              south
                                 30 f
      ---------r1----------
     |-----r8--------|--r1-|     this is better because the small cube
     |               |--r8-|     can be a lot warmer than the room, so
     |               r 400 |     it can store more heat than the same
     r               8  lb r     thermal mass at room temperature, and
     8         t?     --r8-8     the room temperature can be controlled 
     |                     |     by varying the heat flow from the cube
     |                     |     to the room. we can even use a room
     |                     |     temperature thermostat! how modern!
      ---------r8----------

so now we have two equations and steady-state state temperatures,
one for the cube and one for the room.

for the small cube with temperature tc, 

     4k btu = 6h(tc-t)4ft^2/r1      for the south wall, during the day
           + 18h(tc-30)4ft^2/r9     for the south wall, at night
           + 24h(tc-t)5x4ft^2/r8    for the other cube walls,

            = 92tc - 64t - 240,  so tc = 0.696t + 46.1.

for the room with temperature t,

    60k btu = 6h(t-30)60ft^2/r1     daytime south wall loss 
            - 6h(tc-t)4ft^2/r1      daytime gain from small cube glazing
           + 18h(t-30)60ft^2/r9     nighttime south wall loss
           - 24h(tc-t)5x4ft^2/r8    gain from other small cube walls
           + 24h(t-30)5x64ft^2/r8   loss from other room walls,
 
	    = 1524t - 84tc - 43.2k  = 1524t - 84(0.696t + 46.1) -43.2k,

and  106.9k = 1466t, so t = 72.9 and tc = 96.9 f, if i did that right.

the room still has a cloudy day thermal conductance of about 47 btu/f,
so it needs 24h(70-30)47 = 45k btu to stay warm on a 30 f cloudy day.
the cube might keep it warm for about 400(96.9-70)/45k = 0.24 days.
not enough.

let's try doubling the room insulation. now
                       
    60k btu = 6h(t-30)60ft^2/r1     daytime south wall loss 
            - 6h(tc-t)4ft^2/r1      daytime gain from small cube glazing
           + 18h(t-30)60ft^2/r17    nighttime south wall loss
           - 24h(tc-t)5x4ft^2/r16   gain from other small cube walls
           + 24h(t-30)5x64ft^2/r16  loss from other room walls,
 
	    = 968t - 54tc - 27.1k  = 958t - 54(0.696t + 46.1) - 27.1k,

so    89.6k = 920t, so t = 97.4 f and tc = 113.9 f, and the house needs
23k btu on a cloudy day, so the small cube can supply heat for about
400(113.9-70)/23k = 0.76 days. still not enough. but keeping the house
at 101.4 f on average days is uncomfortable and wastes solar energy.

keeping the house at 70 makes the sunspace and small cube warmer, so the
small cube can store heat for longer periods. let's try that, and change
the cube to a 2x4x8' tall closet with about 3,200 pounds of water inside,
and let the sunspace help keep the room warm for 6 hours a day, when it
needs 6h(70-30)5x64ft^2/r16 = 4,800 btu, and let the closet supply the
rest of the 18h(70-30)6x64ft^2/r16 = 17.3k btu per day. with a sunspace
temperature ts, from the closet's point of view 

    32k btu = 6h(tc-ts)32ft^2/r1      day loss from south wall
           + 18h(tc-30)32ft^2/r9      night loss from south wall
           +  6h(tc-70)80ft^2/r8      supplied to room during the day
           + 17.3k                    supplied to room at night

so   20.8k  = 316tc - 192ts, and      tc = 0.608ts + 65.8.

from the sunspace's point of view

    32k btu = 6h(ts-30)64ft^2/r1    daytime south wall loss 
            - 6h(tc-ts)16ft^2/r1    daytime gain from closet glazing
            + 3,200                 supplied to room during the day
	    - 6h(tc-70)80ft^2/r8    supplied to room by closet
 
   so 36.1k = 480ts - 156(0.608ts + 65.8) = 385ts - 10.3k, and
      46.4k = 385ts, and ts = 120 and tc = 139.

the house still needs 23k btu on a cloudy day, but the warmer closet
with more water can keep the house at exactly 70 f for about
3200(139-80)/23k = 9.6 cloudy 30 f days.

on a really cold night (ashrae says it's warmer than -10 f in phila
99% of the time), the closet needs to supply room heat at a rate of
(70-(-10))6x64ft^2/r16 = 1920 btu/h. if closet heat distribution is
convective (vs a fan), using, say, a 2x2' foamboard damper at the top
of the closet, with a $50 6161b12000 honeywell damper motor that uses
2 watts when moving, with a 2x2' return grille at the bottom, we need
1920 = cfm dt = 16.6 4ft^2 sqrt(6')dt^1.5, according to one empirical
chimney formula, which makes the minimum closet-to-room-temperature
difference dt = 5.2 f. that reduces the closet's useful stored heat
to 3200(139-75.2) and its cloudy day lifetime to 8.9 days.

we might use another tiny motorized damper to allow warm sunspace air
to heat the room on an average day, when dt = 50 f and the room only
needs 800 btu/h, so ignoring the heat loss from the inside walls of
the closet, 800 = cfm dt = 16.6 av sqrt(8') 50f^1.5, for a minimum
upper damper area av = 0.05 ft^2, eg a 1' x 1" slot. we might want to
increase the closet glazing area to decrease the sunspace temperature
in this case.

the closet probably needs a fan to circulate hot air between its small
water containers and air heater. storing 17k btu/day over 6 hours means
a heat transfer rate of about 3k btu/h. with a 10 f airflow temperature
difference, this means about 300 cfm, which might come from a $70 10"
diameter 560 cfm 4c688 grainger 36 watt equipment cooling fan with a
temperature rating of 149 f. 

this system is not entirely passive, but the fan might only consume about
36wx6h = 216 watt-hours or 0.2 kwh/day, and if the two dampers only move
for say, 5 seconds every 10 minutes, they only add 4wx5/600x24h = 0.8
watt-hours. this system supplies about 22k btu or 6.5 kwh/day of useful
heat, which makes the cop (the ratio of heat energy moved to electrical
energy needed to move it) 6.5/0.217 = 29.8, dissapointingly low for a
solar heating system, but 10x more than a typical heat pump's cop of 3.

the culprit is that fan... with a foamboard damper that swings open at
the top and another that swings open at the bottom, the closet might only
consume about 2wx2x5sec/3600s/h = 0.006 watt-hours per day, raising the
cop to 6.5/0.000806 = 8,065, and lowering the noise level. for that, we
need a smooth airpath through the closet and 300 = 16.6avsqrt(8'x10f),
so av = 2ft^2, eg a 6" x 4' slot at the top and bottom.

nick

nicholson l. pine                      system design and consulting
pine associates, ltd.                                (610) 489-0545 
821 collegeville road                           fax: (610) 489-7057
collegeville, pa 19426                     email: nick@ece.vill.edyou

computer simulation and modeling. high performance, low cost, solar heating
and cogeneration system design. bsee, msee. senior member, ieee. registered
us patent agent, certified wildland firefighter, grand awards judge for the
50th annual 1999 international science and engineering fair (in philadelphia
this may 5, sponsored by intel, with about 1,500 high school students from
around the world and a few million dollars in scholarships and prizes),
occasional penzance policeman, and 16th century halberdier in a nyc opera
production of gounod's faust. web site: http://www.ece.vill.edu/~nick 

notes:

1. re: "movable insulation," you might say, "oh, i can just pull the
   curtains or thermal drapes or something over all my south windows
   in my passive solar house at night," but curtains and thermal drapes
   are poor insulators, compared to an insulated wall, especially with
   the inevitable air leaks at the edge seals, and people quickly tire
   of operating them manually twice a day. automatic "or somethings"
   can be expensive, and tend to have air leaks too. one might squeeze
   foamboard into the window frames all winter, as a friend does, with
   a dark color and an air gap and a slot at the top for the south window
   foamboard to make solar collectors, but it's more enjoyable to be able
   to see out of the windows during the day (most people don't have much
   desire to look out of windows at night) and use the space between the
   insulation and the windows during the day, no? 

2. tiny cold soap bubbles between two layers of glazing are an exciting
   form of movable insulation. they can make direct gain houses perform.
   they are almost as good as fiberglass, and a lot simpler to move and
   occupy a lot less space than the polystyrene beads used in zomeworks
   beadwalls (tm), a now-unsupported product. bubblewall insulation is
   just starting to take off in greenhouses, after about 30 years of
   kicking the tires of this technology. real houses may soon follow.  

3. a larger building can have a higher cop and a closet that occupies
   a smaller fraction of its floorspace, or one that lives in the
   less expensive sunspace, completely surrounded by insulation. 



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